<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">State a sequence of transformations that will transform the curve with equation $y={{x}^{2}}$on to the curve with equation $y=\frac{1}{4}{{\left( x-3 \right)}^{2}}$.</span></div><p style="text-align: right;">[2]</p></li></ol>
<p>A curve has equation $y=\mathrm{f}\left( x \right)$, where</p>
<p style="text-align: center">$\mathrm{f}\left( x \right)=\left\{ \begin{matrix}
1\mathrm{ for }0\le x\le 1, \\
\frac{1}{4}{{\left( x-3 \right)}^{2}}\mathrm{ for }1&lt;x\le 3, \\
0\mathrm{ otherwise}\mathrm{.} \\
\end{matrix} \right.$</p>
<ol class="i-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Sketch the curve for $-1\le x\le 4$.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">On a separate diagram, sketch the curve with equation $y=1+\mathrm{f}\left( \frac{1}{2}x \right)$, for $-1\le x\le 4$.</span></div><p style="text-align: right;">[2]</p></li></ol>

State a sequence of transformations that will transform the curve with equation $y={{x}^{2}}$on to the curve with equation $y=\frac{1}{4}{{\left( x-3 \right)}^{2}}$.

A curve has equation $y=\mathrm{f}\left( x \right)$, where

$\mathrm{f}\left( x \right)=\left\{ \begin{matrix}
1\mathrm{ for }0\le x\le 1, \\
\frac{1}{4}{{\left( x-3 \right)}^{2}}\mathrm{ for }1&lt;x\le 3, \\
0\mathrm{ otherwise}\mathrm{.} \\
\end{matrix} \right.$

On a separate diagram, sketch the curve with equation $y=1+\mathrm{f}\left( \frac{1}{2}x \right)$, for $-1\le x\le 4$.

Modulus Functions

Functions

International Baccalaureate Higher Level Mathematics

IB HL Math

N2015 P1 Q5

Transformation I: Translation of $3$ units in the positive $x$-direction (replace $x$ with $x - 3$).

Transformation II: Scaling with a factor of $\dfrac{1}{4}$ parallel to the $y$-axis (replace $y$ with $4y$).

$y = (x - 3)^2 \longrightarrow y = \dfrac{1}{4}(x - 3)^2$

For $-1 \leq x < 0$: $y = 0$ (on the $x$-axis).

For $0 \leq x \leq 1$: $y = 1$ (horizontal line).

For $1 < x \leq 3$: $y = \dfrac{1}{4}(x - 3)^2$ (parabola decreasing from $(1, 1)$ to $(3, 0)$).

Key points: open circle at $(0, 0)$, closed circles at $(0, 1)$, $(1, 1)$, $(3, 0)$.

Transformation I: Scaling with a factor of $2$ parallel to the $x$-axis (replace $x$ with $\dfrac{1}{2}x$).

Transformation II: Translation of $1$ unit in the positive $y$-direction (replace $y$ with $y - 1$).

For $-1 \leq x < 0$: $y = 1$. For $0 \leq x \leq 2$: $y = 2$. For $2 < x \leq 4$: $y = 1 + \dfrac{1}{4}\left(\dfrac{x}{2} - 3\right)^2$.

At $x = 4$: $y = 1 + \dfrac{1}{4}(2 - 3)^2 = \dfrac{5}{4}$. Key point: $\left(4, \dfrac{5}{4}\right)$.

N2015 P1 Q5

N2015 P1 Q5

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